Question
Suppose p , q , r 0 and system of equation ( p + a ) x + by + cz =0 a x+(q+b) y+c z=0 a x+b y+(r+c) z=0 has a non-trivial solution, then value of a p + b q + c r is
BITSAT
Mathematics
Determinants
2020
BITSAT 2020
BITSAT Mathematics Question (2020) — Solution
Options
- A. -1
- B. 0
- C. 1
- D. 2
Answer
A. -1
Step-by-step solution
Since the system is homogeneous and has a non-trivial solution 0= | array ccc p+a & b & c \\ a & q+b & c \\ a & b & r+c array |= | array ccc p & -q & 0 \\ 0 & q & -r \\ a & b & r+c array | using R _ 1 R _ 1 - R _ 2 and R _ 2 R _ 2 - R _ 3 Expanding along C , we get p [ q ( r + c )+ br ]+ aqr =0 pqr + pqc + prb + qra =0 a p + b q + c r =-1
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